Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int L[100009], T[100009];
struct Edge {
  int v, next;
} edge[200009];
vector<int> G[100009], M[100009][20];
int P[100009][20];
int n, m, q, tot, head[100009];
inline void addedge(int u, int v) {
  edge[tot].v = v;
  edge[tot].next = head[u];
  head[u] = tot++;
}
void dfs(int u, int fa, int dep) {
  L[u] = dep;
  T[u] = fa;
  int v;
  for (int e = head[u]; e != -1; e = edge[e].next) {
    v = edge[e].v;
    if (v == fa) continue;
    dfs(v, u, dep + 1);
  }
}
inline void merge(vector<int>& ans, vector<int>& b, vector<int>& c) {
  vector<int> a;
  int p = 0, q = 0, szp = b.size(), szq = c.size();
  while ((int)a.size() < 10 && (p < szp || q < szq)) {
    int val;
    if (p < szp && q < szq) {
      if (b[p] < c[q])
        val = b[p], p++;
      else
        val = c[q], q++;
    } else if (p < szp)
      val = b[p], p++;
    else
      val = c[q], q++;
    a.push_back(val);
  }
  ans = a;
  return;
}
void init() {
  dfs(1, 0, 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < (int)G[i].size() && j < 10; j++)
      M[i][0].push_back(G[i][j]);
    P[i][0] = i;
  }
  for (int j = 1; j < 20; j++) {
    for (int i = 1; i <= n; i++) {
      int fa = T[P[i][j - 1]];
      P[i][j] = P[fa][j - 1];
      merge(M[i][j], M[i][j - 1], M[fa][j - 1]);
    }
  }
}
vector<int> solve(int p, int q) {
  vector<int> ans;
  if (L[p] < L[q]) swap(p, q);
  if (L[p] > L[q]) {
    for (int i = 19; i >= 0; i--) {
      if (L[P[p][i]] - 1 >= L[q]) {
        merge(ans, ans, M[p][i]);
        p = T[P[p][i]];
      }
    }
  }
  if (p == q) {
    merge(ans, ans, M[p][0]);
    return ans;
  }
  for (int i = 19; i >= 0; i--) {
    if (P[p][i] != P[q][i]) {
      merge(ans, ans, M[p][i]);
      merge(ans, ans, M[q][i]);
      p = T[P[p][i]];
      q = T[P[q][i]];
      if (p == q) break;
    }
  }
  merge(ans, ans, M[p][0]);
  return ans;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  memset(head, -1, sizeof(head));
  tot = 0;
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    G[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) sort(G[i].begin(), G[i].end());
  init();
  vector<int> ans;
  while (q--) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    ans = solve(u, v);
    printf("%d", min((int)ans.size(), a));
    for (int i = 0; i < (int)ans.size() && i < a; i++) printf(" %d", ans[i]);
    puts("");
  }
  return 0;
}